| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Topic | Proof by induction |
| Type | Suggest and prove formula |
| Difficulty | Challenging +1.2 This is a structured induction question with clear scaffolding: compute derivatives to spot a pattern, conjecture the formula, then prove it. The pattern (coefficients involving powers of 2) emerges naturally from the product rule, and the inductive step is straightforward application of differentiation rules. While it requires multiple techniques and careful algebra, it's a standard Further Maths induction exercise without requiring novel insight. |
| Spec | 1.07q Product and quotient rules: differentiation4.01a Mathematical induction: construct proofs |
**(i)** $y'(x) = (2x+1)e^{2x}$, $\quad y''(x) = (4x+4)e^{2x}$, **B1 B1**
$y'''(x) = (8x+12)e^{2x}$, $\quad y^{(4)}(x) = (16x+32)e^{2x}$ **B1 B1**
**[4]**
**(ii)** Conjecture $\dfrac{\mathrm{d}^n y}{\mathrm{d}x^n} = (2^n x + n\cdot 2^{n-1})e^{2x}$ One mark each: coefficient of $x$, constant **B1 B1**
**[2]**
**(iii)** Differentiating their conjectured expression (must be linear $\times e^{2x}$) **M1**
$\dfrac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} = 2\times(2^n x + n\cdot 2^{n-1})e^{2x} + 2^n\times e^{2x}$ **FT** max 1/2 **A1 A1**
$= (2^{n+1}x + (n+1)\cdot 2^{(n+1)-1})e^{2x}$ Shown of correct form **A1**
Usual induction round-up/explanation of proof, including clear demonstration that $(n+1)^{\text{th}}$ formula is in the right form. **E1**
**[5]**12 Given $y = x \mathrm { e } ^ { 2 x }$,\\
(i) find the first four derivatives of $y$ with respect to $x$,\\
(ii) conjecture an expression for $\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } }$ in the form $( a x + b ) \mathrm { e } ^ { 2 x }$, where $a$ and $b$ are functions of $n$,\\
(iii) prove by induction that your result holds for all positive integers $n$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q12 [11]}}